Spring MVC REST 服务和客户端。我正在寻找一种更好的方法来处理错误!
我有时想向客户端返回错误消息或状态代码,但我不知道如何。有人可以告诉我如何找到更好的方法来处理 Spring REST 服务和客户端的错误和错误消息。
这是我的服务代码:
@RequestMapping(value = "/{name}", method = RequestMethod.GET)
@ResponseBody
public User getName(@PathVariable String name, ModelMap model) throws ResourceNotFoundException
{
logger.debug("I am in the controller and got user name: " + name);
/*
Simulate a successful lookup for 2 users, this is where your real lookup code would go
*/
if ("user2".equals(name))
{
return new User("User2 Real Name", name);
}
if ("user1".equals(name))
{
return new User("User1 Real Name", name);
}
throw new ResourceNotFoundException("User Is Not Found");
}
@ExceptionHandler(ResourceNotFoundException.class)
public ModelAndView handleResourceNotFoundException(ResourceNotFoundException ex)
{
logger.warn("user requested a resource which didn't exist", ex);
return new ModelAndView( jsonView, "error", "user requested a resource which didn't exist");
}
现在这里是客户端的代码:
Map<String, String> vars = new HashMap<String, String>();
vars.put("name", "user1");
/**
*
* Doing the REST call and then displaying the data/user object
*
*/
RestTemplate restTemplate = new RestTemplate(commons);
restTemplate.getMessageConverters().add(new MappingJacksonHttpMessageConverter());
restTemplate.getMessageConverters().add(new StringHttpMessageConverter());
try
{
User jsonreturn = restTemplate.getForObject("http://" + mRESTServer.getHost() + ":8080/json/{name}", User.class, vars);
LOGGER.debug("return object: " + jsonreturn.toString());
}
catch(Exception e)
{
LOGGER.error("error: " + e.toString());
}
如果找不到用户并且我的客户端代码收到此错误,我想找到返回状态代码和消息的方法:
org.springframework.http.converter.HttpMessageNotReadableException: Could not read JSON: Unrecognized field "error"